The black knight is our lion chasing 8 white bishop zebras which can't capture. Can the zebras evade the lion forever, if team zebra positions all the pieces and has the first turn?

Rules:

The game is played on a standard 8x8 chess board

Zebras move like bishops but can't capture

Lions move like knights and only need to capture one zebra to win

The Zebra team will always play their best game to avoid capture

Zebras get to position all pieces to their advantage

Zebras can choose to first move if they desire

Each side moves a single piece per turn

Question:
If there are 8 zebras and only 1 lion, can the lion win?

If you lion can win how does it do so? If the zebras always win (escape) where on the board do they and the lion start and what is their strategy?

While the answers so far are a good start they still don't completely answer the question. I might offer a bounty on this question and I would definitely award it to anyone who can prove that the Lion can always win. (if that is the case)

4 Answers
4

Then whenever the knight attacks any piece, I will move it to its adjacent diagonal so that it is contained within these boxes :

Now, the knight cannot fork two pieces inside same box(as they are of opposite colour). Also knight cannot fork two pieces from two boxes as they are quite apart...(4 places between them). So I will just have to make sure that the knight does not capture me in next move and I am safe...

Note that if the knight attacks from c3(only time when I will have to leave box) or such similar square I will diagonally move to the opposite box. From c3 he will have to go to e2 then g3 then f1 then e3 then his attacks will end (any other move will also mean end of attack). Then I will take back my piece to where it was. So initial condition.

$\begingroup$@Lawrence white in piece colour, not square. If it were square then you should be much more confused at the reference to black knight.$\endgroup$
– LeppyR64May 19 '15 at 14:16

2

$\begingroup$In the section detailing c3 how do you guarantee that the spot you are moving your zebra to doesn't contain a zebra already? If you can not guarantee this; have you considered the followup of d5 rather than e2?(blocking the return path of the zebra).$\endgroup$
– TaemyrMay 20 '15 at 0:49

Note: The following full answer expands on the previous partial answer, which has been retained below.

Full answer

To analyze all the possible states, the algorithm Ken Thompson described in his 1986 paper Retrograde Analysis of Certain Endgames, which was used to develop some of the earliest chess endgame tablebases, was adapted for this question. His method starts at known winning positions and moves backward in time ("unmoves") to predecessor positions. Those positions are analyzed and the ones that always lead to defeat are added to the list of losing positions. Then new predecessor positions of the losing positions are analyzed. This continues until there are no new predecessors.

At the end of the analysis, either all possible positions have been added to the list of losing positions and a win is guaranteed with optimal play, or there is a subset of non-losing positions that can be leveraged to defend against a win.

Analysis of Lions and Zebras on a Chess Board leads to the conclusion that...

In the jungle, the mighty jungle, the lion eats tonight...

An optimal lion wins from every possible game position. Against optimal zebras each, the lion slightly improves with each move until his opponents run out of defenses. This takes 16 moves at most.

Lion eats in 16

This is a typical best-case setup for the zebras. They are arrayed in the corners and the lion lacks any immediate threatening moves. But even if the zebras could dictate the lion's first move, it would offer them no advantage. All the lion's initial moves lead to a captured zebra in 16 moves.

Since the state space is too large to present a full analysis for the tens of billions of positions, the following analysis has been limited to this position. Further, it only looks at lines where the lion chooses Nd1 as his first move. It produces a number interesting lines that demonstrate the strategies without having an unmanageable number of variations.

Interestingly, in responding to Nd1, despite a plethora of options, the zebras only have two optimal replies: Bab8 and Bab1. Anything else results in earlier capture.

Analysis of the lines starting at this position are given in PGN format. Given the non-standard rules of the game, not all PGN viewers can handle them. But, as of this writing, the PGN viewer on chess.com works. (Choose Load PGN and paste in the PGN.). Zebra moves are annotated for only optimal move: '!'; suboptimal move: '?'; or highly suboptimal move: '??'.

1... Nd1 2. Bab8

In moves 3 to 11, the zebras maintain their defensive posture by occupying the corners. But they only have one optimal move at each turn. On move 12, the zebras finally two options Bab6 or Bab2, but either one forces the abandoment of the corners. Either way, the lion continues with b4, then c2, and on move 14 the a1 zebra then has four choices: b2, c3, e5, or f6. The lion continues to e3 and then no matter which of the eight combinations of moves the zebras chose on 12 and 14, the lion forks them on move 15.

1... Nd1 2. Bab1

The lion leaves the zebras with only a single optimal choice on moves 3 to 10. Unlike the Bab8 line, this line requires early abandonment of the corners. The zebras are finally afforded a choice on move 11, either Bc8 or Be8. But the table is already set and Nf7 forces Bhb2. The lion then threatens one zebra after another until he finally captures the a1 zebra trapped on move 12.

The amazing thing about this line is that if the zebras play optimally the lion could play blind. In optimal play, the lion can use these same moves starting at move 2 and he will always capture the zebra on a1 on move 16.

Other variations

The following lines show the optimal lion strategy to reach a win in five moves. The starting position is after 1... Nd1. Reviewing with a PGN viewer is recommended since there are many subvariations, and many of those deeply nested lines.

Conclusion

The preceding variations demonstrate the tactics adopted by the lion versus various zebra defenses. It's not possible to present the full analysis, but the final result is the lion always wins.

Partial answer

Note: This was the first cut at the game analysis prior to the full analysis done above.

If we limit the zebras to execute the four-corners strategy suggested in other answers, the problem is more manageable. An analyzing every state where the zebras restrict themselves to the four squares in each corner shows that this strategy is unsustainable. Sooner or later, the zebras have to abandon those squares.

To demonstrate this, I implemented a Python program to with an optimal lion. (It should work in both Python 2.7 and Python 3.x, but let me know if it not.) The user takes the role of the zebras. To reduce the state space, I limited the zebras to having four on white and four on black with two zebras on each long diagonal. This is the optimal organization for the zebras when playing the four corners.

Once the program is started, you receive a command prompt.

To create an initial position, use the start command followed by {l}:{z1}{z2}...{z8} where {l} is the lion's square and {zn} is each zebra's square. For instance, to use the position shown in Prem's answer use:

start d5:a1a2a7a8h1h2h7h8

The lion will move and the next position shown. To move just enter a move for a zebra, which must be to one of the corner squares. For example:

a2b1

The zebra move will be shown, along with the optimal reply from the lion.

$\begingroup$Can you explain what you call optimal (how does the lion react)? Going through code is hard and evrybody is not a master in Python. Thanks for the script however!$\endgroup$
– UntitpoiJan 23 '18 at 8:35

$\begingroup$@Untitpoi By optimal I mean that the lion always plays moves that lead him closer to the capturing a zebra. The bulk of the code is a compressed table that just holds the lion's optimal move for each state. It wasn't really intended to be read, but just to allow people to play the four-corner defense against an optimal lion and show that that strategy does not hold. In a comment on the OP, justhalf asked if someone could analyze the answers given, so that's what this partial answer is.$\endgroup$
– D KruegerJan 23 '18 at 10:02

$\begingroup$So, after 3 years the problem seems to be finally solved.$\endgroup$
– User Not FoundMar 7 '18 at 6:37

$\begingroup$@UserNotFound I tried for some time to find a way to present a complete proof, but failed. It's fascinating how nuanced the game is. Maybe I'll endeavor a little more to reduce the full state space into something that will fit in an answer.$\endgroup$
– D KruegerMar 7 '18 at 12:52

$\begingroup$Is there any simple way to show that the lion wins in all other combinations of light/dark-square zebras? The lion's victory stems from its ability to find ways of moving twice for each light-square zebra move. If there were fewer dark-square zebras, it might not be able to find all the opportunities it might need. I'd guess that the lion would gain more advantage by having extra light-square targets than it would lose by having fewer dark-square ones, but can that be easily proven?$\endgroup$
– supercatMay 14 '18 at 20:26

Place 4 Zebras on White, and 4 Zebras on Black.
At any move, Lion can attack only Black or only White, so only 4 pieces.

Keep the 4 Black Zebras in the 4 corners, or adjacent to the corners.
Keep the 4 White Zebras in the 4 corners, or adjacent to the corners.
Now, only one Zebra can be in danger at any move. So only move that Zebra to the opposite corner.

Conclusion : Zebras can always escape.

In the figure , the blue lines are two possible ways to attack the left top corner Zebras, but the gray lines are the responses to escape. Same applies for all other corners.

Escape Algorithm Overview :
The Lion can be in only one of the 4 big quadrants (left bottom, left top, right bottom, right top) and so it can attack only one of the two Zebras in that quadrant. That Zebra has to move to the opposite corner.
If the Lion also moves to that opposite quadrant, then that Zebra must go back to initial position.
If the Lion moves to any other quadrant, then one Zebra from that quadrant will so move to opposite quadrant.
If a piece may come into danger by moving, then simply move a Zebra of the other colour.

Definition of opposite corner :
For the Zebras, there are 8 corners formed by left/right or top/bottom or inner/outer.
For getting opposite, all 3 Dimensions must change: left must become right and vice-versa , top must become bottom and vice-versa , inner must become outer and vice-versa. Example : top-left-outer corner is opposite of bottom-right-inner corner.

Aim of every move is Either (1) If a piece is in danger, move that piece to opposite corner Or (2) If no piece is in danger, (2A) Move a piece back to original position, provided it is a safe move and the Lion is not blocking the way Or (2B) Move a piece of the opposite colour, so that no piece comes into danger by that move Or (2C) Move a Zebra of the same colour and the same quadrant as the Zebra blocked by the Lion in case (2A).

Explanation of (2C) : When Lion moves from D5 to C3, one Zebra moves from A2 to G8. Now if Lion moves back to D5, Zebra can not use condition (2A) to go back to original position A2, because L is blocking the way. So we have to use condition (2C) to move one Zebra from H7 to B1. Still we are maintaining 2 Zebras of opposite colour in each quadrant.

$\begingroup$When you move the zebra to the other corner you have two zebras on that color in that corner. Can you guarantee that it will be safely able to get back to the original corner?$\endgroup$
– TaemyrMay 19 '15 at 13:02

$\begingroup$@Taemyr , Well, the Lion will take many moves to come to that corner, so in the meantime the Zebras have to come back to the initial positions.$\endgroup$
– PremMay 19 '15 at 13:03

2

$\begingroup$@Prem Starting from your picture, lion moves: C7 B5 C3 E2 G3 H1 killing the zebra on H1 because it is trapped by the zebra at G2.$\endgroup$
– JS1May 20 '15 at 20:24